I have attached an image of the diagram showing the nature of this motion
Answer:
Protons speed = 2.96 x 10^(5) m/s
Explanation:
A) At closest point of approach to the positive plate, the proton came to rest momentarily.
Thus;
Loss in Kinetic Energy = Gain in Electric potential energy
Hence;
(1/2)(mv^(2)) = eΔV
So, ΔV = (mv^(2))/(2e)
Mass of proton = 1.673 × 10-27 kilograms
Proton elementary charge(e) = 1.6 x 10^(-19) coulumbs
And from the question v = 200,000 m/s
So, ΔV = [1.673 × 10^(-27) x 200000^(2)] / (2 x 1.6 x 10^(-19)) = 209 V
This is less than 250V which is half of the charge at the positive plate shown in the diagram.
Therefore, the speed is insufficient to reach the positive plate from P to Q.
B) Gain in KE = qΔV
Thus; 1/2mvf^(2) - 1/2mvi^(2) = eΔV
Where, vf is final velocity and vi is initial velocity.
So simplifying, we get;
vf^(2) - vi^(2) = (2eΔV)/m
So, vf = √[(2eΔV)/m) + (vi^(2))
= √[(2 x 1.6 x 10^(-19) x 250)/(1.673 × 10^(-27)) + (200,000^(2))
= 2.96 x 10^(5) m/s
